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You have a normal 52 card deck. Note that your hand does not depend on order. You are dealt two cards. What is the probability you will get at least one ace?

How do I approach this problem?

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  • $\begingroup$ What's the probability of getting no aces? $\endgroup$ Apr 27, 2017 at 4:23
  • $\begingroup$ @LordSharktheUnknown remaining 48 cards are not aces $\endgroup$
    – socrates
    Apr 27, 2017 at 4:26
  • $\begingroup$ Is it $1-(48/52C2)$? $\endgroup$
    – socrates
    Apr 27, 2017 at 4:27
  • $\begingroup$ How many hands are possible? How many hands with no aces are possible? $\endgroup$
    – shardulc
    Apr 27, 2017 at 4:27
  • $\begingroup$ @shardulc $52C2$; $52C48$ $\endgroup$
    – socrates
    Apr 27, 2017 at 4:29

5 Answers 5

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Instead, calculate the probability that you won't get an ace in two picks.

This you do by saying that there are $48$ non-aces in the deck, so you pick any two of those in $\binom{48}2$ ways, and the ways of picking two cards from the complete deck is $\binom{52}2$. So the probability of not picking an ace, is $\frac{\binom{48}2}{\binom{52}2}$.

The answer you are then looking for is $1 - \frac{\binom{48}2}{\binom{52}2}$.

EDIT : For those who want a nice looking answer,it's $\frac{33}{221}$.

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The denominator is the number of 2 card hands, which is ${52 \choose 2}$. The numerator is the number of 2 card hands which contain at least one ace. This quantity is ${52 \choose 2}$ minus the number of 2 card hands which do not contain any aces, and hence is equal to ${52 \choose 2} - {48 \choose 2}$. Thus, the probability in question is $\frac{{52 \choose 2}-{48 \choose 2}}{{52 \choose 2}}$.

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Let $A$ be the event of getting at least one ace, thus $A'$ be defined as the complement event which is getting no ace. $P(A) = 1 - P(A') = 1 - \dfrac{\binom{48}{2}}{\binom{52}{2}}=...$

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Let A be the event of getting at least one ace

Let A1 be the event of getting ace in 1st card and non-ace in 2nd card

Let A2 be the event of getting non-ace in 1st card and ace in 2nd card

Let A3 be the event of getting ace both 1st card and 2nd cards

Thus P(A) = P(A1) + P(A2) + P(A3)

P(A1) = 4/52 * 48/51

P(A2) = 48/52 * 4/51

P(A3) = 4/52 * 3/51

P(A) = (4 * 48 + 4 * 48 + 4 * 3)/(52 * 51) = 33/221

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If this can help you approach the problem in a simpler way:

you can have 4 cases:

  1. Ace, Not Ace
  2. Not Ace, Ace
  3. Ace, Ace
  4. Not Ace, Not Ace

To get AT LEAST 1 ace means you have to sum the probabilities of first 3 cases.

First case: `4/52*48/51` = 0.07240
Second case: is the same as the first => 0.07240
Third case: `4/52*3/51` = 0.00452

Sum of the three cases is: 0.14932

Approach:

At the denominator you notice that it's always 52*51 which come from the permutation formula n!/(n-r)! where n=52 and r=2. This number represents all the possible permutations of 2 cards from a deck of 52.

At the numerator you need to calculate the subset of the total permutations above that should be considered as "success" cases.

  • The first case says that you can have 4 aces in the first card and 48 'not aces' in the second. By the basic principle of counting there are 4*48 different ways to have this event.
  • The second case follow the same reasoning approach.
  • The third case says that you can have 4 aces in the first card and 3 left possible aces in the second card. So you have 12 possible permutations of aces.

Add the first 3 success cases and divide for the overall permutations to have the probability of at least one ace.

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